Question: Complete the square to solve for $x$. $x^{2}-6x = 0$
Explanation: We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-6$ , half of it would be $-3$ , and squaring it gives us ${9}$ $x^2 - 6x { + 9} = 0 { + 9}$ We can now rewrite the left side of the equation as a squared term. $( x - 3 )^2 = 9$ Take the square root of both sides. $x - 3 = \pm3$ Isolate $x$ to find the solution(s). $x = 3\pm3$ So the solutions are: $x = 6 \text{ or } x = 0$ We already found the completed square: $( x - 3 )^2 = 9$